The Binomial Theorem

Lecture 29

Section 6.7

Mon, Apr 3, 2006

The Binomial Theorem

Theorem: Given any numbers a and b and any nonnegative integer n,

(a + b)n = k = 0..n C(n, k)an – kbk. Proof: Use induction on n. Basic step: Let n = 0. Then

(a + b)0 = 1. k=0..0 C(0, k)a0 – kbk = C(0, 0)a0b0 = 1.Therefore, the statement is true when n = 0.

Proof, continued

Inductive stepSuppose the statement is true for some n

0.Then

n

k

kknn

k

kkn

n

k

kkn

nn

bak

nba

k

n

bak

nba

bababa

0

1

0

1

0

1

Proof, continued

n

k

nkknn

n

k

n

k

nkknkknn

n

k

n

k

nkknkknn

bbak

n

k

na

bbak

nba

k

na

bbak

nba

k

na

1

111

1 1

1111

1

1

0

1111

1

1

Proof, continued

Therefore, the statement is true for n + 1. Thus, the statement is true for all n 0.

.1

1

1

0

1

1

111

n

k

kkn

n

k

nkknn

bak

n

bbak

na

Example: Binomial Theorem

Expand (a + b)8.C(8, 0) = C(8, 8) = 1.C(8, 1) = C(8, 7) = 8.C(8, 2) = C(8, 6) = 28.C(8, 3) = C(8, 5) = 56.C(8, 4) = 70.

Example: Binomial Theorem

Therefore,

(a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3

+ 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8.

Example: Calculating 1.016

Compute 1.018 on a calculator. What do you see?

Example: Calculating 1.016

1.018 = (1 + 0.01)8

= 1 + 8(0.01) + 28(0.01)2 + 56(0.01)3

+ 70(0.01)4 + 56(0.01)5 + + 28(0.01)6 + 8(0.01)7 + (0.01)8

= 1 + .08 + .0028 + .000056 + .00000070

+ .0000000056 + .000000000028 + + .00000000000008

+ .0000000000000001

= 1.0828567056280801.

Example: Approximating (1+x)n

Theorem: For small values of x,

and so on.

.6

)2)(1(

2

)1(11

.2

)1(11

.11

32

2

xnnn

xnn

nxx

xnn

nxx

nxx

n

n

n

Example

For example,

(1 + x)8 1 + 8x + 28x2

when x is small. Compute the value of (1 + x)8 and the

approximation when x = .03.

Newton’s Generalization of the Binomial Theorem

Theorem: Given any numbers a, b, and n,

(a + b)n = k=0.. C(n, k)an – kbk.

where

C(n, k) = [n(n – 1)…(n – k + 1)]/k! Note that n need not be an integer nor

positive.

Newton’s Generalization

Expand (a + b)-1 in a series, showing the first 5 terms.

Compute the first 5 coefficientsC(-1, 0) = 1.C(-1, 1) = -1.C(-1, 2) = (-1)(-2)/2! = 1.C(-1, 3) = (-1)(-2)(-3)/3! = -1.C(-1, 4) = (-1)(-2)(-3)(-4)/4! = 1.

Newton’s Generalization

Therefore,

...1

...1

...

4321

44332211

453423211

a

b

a

b

a

b

a

ba

babababaa

babababaaba

Newton’s Generalization

Expand (a + b)5/2 in a series, showing the first 5 terms.

.8

15

!223

25

2,2

5

.2

51,

2

5

.10,2

5

C

C

C

.128

5

384

15

!421

21

23

25

4,2

5

.16

5

48

15

!321

23

25

3,2

5

C

C

Newton’s Generalization

Therefore,

...128

5

16

5

8

15

2

51

...128

5

16

5

8

15

2

51

...128

5

16

5

8

15

2

5

4322/5

44332212/5

42/332/122/12/32/52/5

a

b

a

b

a

b

a

ba

babababaa

babababaaba

Newton’s Generalization

If 0 < b < a, then this series will converge rapidly.

Newton’s Generalization

Approximate (1.2)5/2. Let a = 1 and b = 0.2 = 1/5. Then b/a = 1/5.

.5774375.116000

2523916000

1

400

1

40

3

2

11

...5

1

128

5

5

1

16

5

5

1

8

15

5

1

2

512.1

4322/5

The Multinomial Theorem

Theorem: In the expansion of

(a1 + … + ak)n,

the coefficient of a1n1a2

n2…aknk is

n!/(n1!n2!…nk!)

where n = n1 + n2 + … + nk.

Example: The Multinomial Theorem

Expand (a + b + c + d)3. The terms are

a3, b3, c3, d3, with coefficient 3!/3! = 1.a2b, a2c, a2d, ab2, b2c, b2d, ac2, bc2, c2d,

ad2, bd2, cd2, with coefficient 3!/(1!2!) = 3.abc, abd, acd, bcd, with coefficient 3!/(1!1!

1!) = 6.

Example: The Multinomial Theorem

Therefore,

(a + b + c + d)3 = a3 + b3 + c3 + d3 + 3a2b

+ 3a2c + 3a2d + 3ab2 + 3b2c + 3b2d

+ 3ac2 + 3bc2 + 3c2d + 3ad2 + 3bd2

+ 3cd2 + 6abc + 6abd + 6acd + 6bcd. Find (a + b + c)4.

Actuary Exam Problem

If we expand the expression

(a + 2b + 3c)4,

what will be the sum of the coefficients?